The question is given from ncert Book of class 10th Ch. no. 5 Ex. 5.2 Q. 11. In the following question you have to find the term which will be 132 more than the 54th term of the given A.P. Give the solution.
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Solution:-
It is given in this A.P. that 3, 15, 27, 39, …
a = 3
d = a2 − a1 = 15 − 3 = 12
We know:-
an = a+(n−1)
Therefore,
a54 = a+(54−1)d
3+(53)(12)
3+636 = 639
a54 = 639
We have to find the term of this series or A.P which is 132 more than a54 Therefore 771.
Let the nth term be 771.
an = a+(n−1)d
771 = 3+(n −1)12
768 = (n−1)12
(n −1) = 64
n = 65
Hence, the 65 term was 132 more than 54th term.